Finding the branch points of $\log(z^2-1)$ This example is presented in the following paper (http://math.mit.edu/classes/18.305/Notes/n00Branch_Points_B_Cuts.pdf, example 2.2, page 11). The author uses the identity $\log(z^2-1)=\log(z+1)+\log(z-1)$ and then proceeds to study the branch points of the two logarithms. However, aren't all these functions multivalued? How does he know that $\log(z+1)$ is continuous on the circle centered at $1$ while $\log(z-1)$ is not? I just cannot make sense of that argument. 
Any help would be appreciated.
 A: The branch points of $f_1(z)=\log(z-1)$ are $A_1=\{1,\infty\}$.  
The branch points of $f_2(z)=\log(z+1)$ are $A_2=\{-1,\infty\}$.  
The branch points of $f(z)=f_1(z)+f_2(z)=\log(z^2-1)$ are $A_1\cup A_2 = \{1,-1,\infty\}$.  
A branch cut for $f_1(z)$ must connect the two points of $A_1$ and a branch cut of $f_2(z)$ must connect the two points of $A_2$.  
To select branch cuts for $f(z)$, we can take two branch cuts, one from each of these points to the point at $\infty$.
The function $f_1(z)$ has a branch point at $z=1$, so any circle surrounding this point will necessarily run into a branch cut.  The function $f_2(z)$ does not have $z=1$ a branch point, so for this function, surrounding the point $z=1$ by a circle poses not difficulty for circles of radius less than 2 (the distance between the points $\{1,-1\}$, as long as we don't run the branch cut through this circle.
UPDATE
The point at $\infty$ is a branch point of $f(z)= f_1+f_2$ and thus the segment $[-1,1]$ is not a valid branch cut.
Thanks to @Maxim for pointing this out in the comments.
